Given GE=a,HF=b,EF=c, and ABCD is cyclic so A+C=B+D=180 and angleEBC=angleEDA=D,angleECB=angleEAD=A=angleFCD,angleFDC=angleFBA=Band angleBCD=angleECF=C=180-A, we have by sine rulefor triangles CB/EC=Sin(A+D)/SinD, CD/CF=Sin(A+B)/SinB=-Sin(A-D)/SinD, so CB/EC + CD/CF=(2CosA.SinD)/SinD=2CosA so CB.CF+CD.CE=2EC.CFCos(180-C) and b^2-FC^2+a^2-CE^2=-2EC.CFCosC hence a^2+b^2=FC^2+EC^2-2EC.CF.CosC so a^2+b^2=c^2

http://img9.imageshack.us/img9/6039/ugqa.pngLet circumcircle of triangle CDF cut EF at MSince ABCD and CDFM are cyclic quadrilaterals It easy to show that angle(BCM) supplement to angle (BEM) => quadrilateral BCME is cyclic.Since E is located on radical line or circles O and CDFM => a^2=EC.ED=EM.EF Since F is located on radical line or circles O and BCME=> b^2=FC.FB=FM.FEAdd above expressions side by side we have a^2+b^2= c^2